Noetherian

Noetherian Rings

Satisfy the ACC on ideals.

Examples

attachments/Pasted%20image%2020220420181700.png

${\mathcal{O}}_{X, x}$ for ${\mathcal{O}}_X({-}) \coloneqq\mathop{\mathrm{Hol}}({-}; {\mathbf{C}})$ the sheaf of holomorphic functions, the stalks are rings of germs of holomorphic functions. These are regular local rings, and Noetherian by the Weierstrass preparation theorem.

Non-Noetherian Rings

$R[x_0, x_1, \cdots]$, a polynomial ring on countably many variables.
For $X = {\mathbf{C}}$ equipped with the sheaf ${\mathcal{F}}= C^\infty({-}; {\mathbf{R}})$, the stalk ${\mathcal{F}}_0$ is the ring of germs of smooth functions at the origin. Take $f(x) \coloneqq e^{-{1\over x^2}}$ and patch $f(0) = 0$ to get a smooth-nonanaltic function, contradicting Krull’s intersection theorem.
$C^0([0, 1]; {\mathbf{R}})$ the ring of continuous functions.
${\mathcal{O}}_K$ for $K = { {\mathbf{C}}_p }= \widehat{ \operatorname{cl}^{\mathrm{alg}} ({\mathbf{Q}}_p)}$

Exercises

Show that Noetherian integral domains have factorization into finitely many irreducibles.
Show that subrings of Noetherian rings need not be Notherian.
Show that subalgebras of finitely generated algebras need not be finitely generated.
Show that if $R$ is Noetherian, then the nilradical ${\sqrt{0_{R}} }^n = 0$ for some $n$.
Show that the Krull’s intersection theorem fails for non-Noetherian rings.
Find a Noetherian ring where each ideal is finitely generated, but the number of generators needed is not uniformly bounded.
Show that in a Noetherian ring, the only element in the intersection of all powers of all maximal ideals is zero.
Show that a subring of a Noetherian ring need not be Noetherian.
Does Noetherian imply Artinian? Or vice-versa?
- What is a sufficient condition to guarantee Noetherian $\iff$ Artinian?
Prove several equivalent characterizations of Noetherian rings.
Show that the homomorphic image of a Noetherian ring is Noetherian.
If $R$ is a Noetherian ring, show that any finitely generated module $M\in {}_{R}{\mathsf{Mod}} ^{\mathrm{fg}}$ is Noetherian.
Show that if $R$ is Noetherian, any Unsorted/localization of rings is Noetherian.
Show that if $R$ is Noetherian then $R[x]$ is Noetherian.
Show that every $A\in \mathsf{Alg} ^{\mathrm{fg}}_{/ {k}} $ for $k\in \mathsf{Field}$ is Noetherian.
Show that if $A \in \mathsf{Alg} ^{\mathrm{fg}}{/ {k}} $ for $k\in \mathsf{Field}$, then if $A$ is additionally a field then $A{/ {k}} $ is a finite algebraic extension of fields.
Show that ${\sqrt{0_{R}} }$ is nilpotent when $R$ is Noetherian.
Show that localizations preserve exactness and being Noetherian.
Show that the ${\mathfrak{m}}{\hbox{-}}$adic completion of a Noetherian local ring is again local.
Show that every ideal in a Noetherian ring has finite height, but may not be finite depth.
- Show that ideals in a local Noetherian ring are finite depth.
Show that Noetherian rings are finite dimensional.
Show that a graded ring $R$ is Noetherian iff $R_0$ is Noetherian and $R$ is a finitely generated $R_0{\hbox{-}}$algebra.

Links to this page

unresolved links output

Artinian in Unsorted/Noetherian

Weierstrass preparation theorem in Unsorted/Noetherian

algebras in Unsorted/HH, -Unsorted/Noetherian

graded ring in Projects/2022 Advanced Qual Projects/Commutative Algebra/010 Definitions, -Unsorted/Noetherian

integral domains in Unsorted/Noetherian

local ring in Projects/2022 Advanced Qual Projects/Commutative Algebra/020 Problems and Exercises, -Unsorted/Formal group, -Unsorted/Hensel’s Lemma, -Unsorted/Noetherian, -Unsorted/scheme

nilradical in Unsorted/Noetherian
order

An order ${\mathcal{O}}$ is a Noetherian integral domain of dimension one with nonzero conductor. Equivalently, \begin{align*} \operatorname{cl}^{\mathrm{int}} {\mathcal{O}}\in {}_{{\mathcal{O}}}{\mathsf{Mod}} ^{\mathrm{fg}}.\end{align*}

Setup: take $R$ a Noetherian integral domain with $K\coloneqq\operatorname{ff}(R)$, and let $\Lambda\in \mathsf{Alg} _{/ {K}} ^{\mathrm{fd}}$. Then an $R{\hbox{-}}$order in $\Lambda$ is a ring $X$ with the structure of an $R{\hbox{-}}$lattice, so $X\leq \Lambda \in {}_{R}{\mathsf{Mod}} ^{\mathrm{fg}}$ such that $\mathop{\mathrm{span}}_K X =\Lambda$, i.e. $X$ is an $R{\hbox{-}}$submodule of $\Lambda$ which spans it as a $K{\hbox{-}}$vector space.
adic completion

Show that the completion of a Noetherian ring $R$ is flat as an $R{\hbox{-}}$module.